如何在laravel查詢中獲取陣列

這是我的查詢

$bbaplaukts = DB::table('materials')
            ->where('statuss', '=', 'Izlietots')
            ->orwhere('statuss', '=', 'Pasutits')
            ->leftJoin('bbaplaukts', 'materials.scanner_code', '=', 'bbaplaukts.scanner_code')
            ->leftJoin('alternatives', 'materials.id', '=', 'alternatives.material_id')
            ->select(
              'materials.scanner_code as scanner_code',
              'materials.reference as reference',
              'materials.description as description',
              'materials.id as id',
              'bbaplaukts.place as place',
              'alternatives.alternatives as alternatives',
              'alternatives.links as links',
              DB::raw('count(IF(statuss = "Izlietots",bbaplaukts.statuss,null)) as izlietots'),
              DB::raw('count(IF(statuss = "Pasutits",bbaplaukts.statuss,null)) as pasutits'))
            ->groupBy('materials.scanner_code', 'materials.reference', 'materials.description',
             'materials.id', 'bbaplaukts.place', 'alternatives.alternatives', 'alternatives.links')
            ->orderBy('bbaplaukts.place', 'asc')
            ->paginate(15);

這是它回傳的結果

#items: array:4 [▼
      0 => {#1303 ▼
         "scanner_code": "DM DS D2 D0"
         "reference": "Metāla ??ēres 20cm"
         "description": "Metāla ??ēres 20cm"
         "id": 2
         "place": "Bbaskapis"
         "alternatives": "Test123"
         "links": "Test123"
         "izlietots": 2
         "pasutits": 0
      }
      1 => {#1322 ▼
         "scanner_code": "DM DS D2 D0"
         "reference": "Metāla ??ēres 20cm"
         "description": "Metāla ??ēres 20cm"
         "id": 2
         "place": "Bbaskapis"
         "alternatives": "Test12345"
         "links": "https://www.google.com/search?client=firefox-b-d&q=talsu riepas"
         "izlietots": 2
         "pasutits": 0
      }

如您所見，它們幾乎相同，只有“替代方案”和“鏈接”不同。有沒有辦法讓查詢回傳單個條目，但“替代”和“鏈接”是所有不同條目的陣列？

對不起，標題不好，真的想不出如何表達這一點。

uj5u.com熱心網友回復：

Laravel 通過Model Relationships提供了一個最有效的特性來實作這一點。但是，如果您更喜歡使用查詢構建器，您可以嘗試從主查詢中洗掉“alternatives”連接并單獨進行：

$bbaplaukts = DB::table('materials')
            ->where('statuss', '=', 'Izlietots')
            ->orwhere('statuss', '=', 'Pasutits')
            ->leftJoin('bbaplaukts', 'materials.scanner_code', '=', 'bbaplaukts.scanner_code')
            ->select(
              'materials.scanner_code as scanner_code',
              'materials.reference as reference',
              'materials.description as description',
              'materials.id as id',
              'bbaplaukts.place as place',
              DB::raw('count(IF(statuss = "Izlietots",bbaplaukts.statuss,null)) as izlietots'),
              DB::raw('count(IF(statuss = "Pasutits",bbaplaukts.statuss,null)) as pasutits'))
            ->groupBy('materials.scanner_code', 'materials.reference', 'materials.description',
             'materials.id', 'bbaplaukts.place')
            ->orderBy('bbaplaukts.place', 'asc')
            ->paginate(15);
            
foreach ($bbaplaukts as $key=>$bbaplaukt){
            $alternatives = DB::table('alternatives')->select('alternatives')->where('material_id', $bbaplaukt->id)->get();
            $links = DB::table('alternatives')->select('links')->where('material_id', $bbaplaukt->id)->get();
            $bbaplaukts[$key]->alternatives = $alternatives;
            $bbaplaukts[$key]->links = $links;
        }
return  response($bbaplaukts, 200);

標籤：mysql 拉拉维尔

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